3.8 Q-13

Question Statement

Solve the differential equation:

$$\operatorname{Sec}^{2} \mathrm{x} \tan \mathrm{y} , dx + \operatorname{Sec}^{2} \mathrm{y} \tan \mathrm{x} , dy = 0$$

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Background and Explanation

This is a first-order, separable differential equation. The goal is to rearrange the equation so that all terms involving $x$ are on one side and all terms involving $y$ are on the other. Once separated, we can integrate both sides to find the solution.

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Solution

1. Start with the given equation:

   The equation is:

$$\operatorname{Sec}^{2} \mathrm{x} \tan \mathrm{y} , dx + \operatorname{Sec}^{2} \mathrm{y} \tan \mathrm{x} , dy = 0$$

2. Rearrange the terms:

   Move the second term to the right-hand side:

$$\operatorname{Sec}^{2} \mathrm{y} \tan \mathrm{x} , dy = -\operatorname{Sec}^{2} \mathrm{x} \tan \mathrm{y} , dx$$

3. Separate the variables:

   Now, divide both sides by $\tan \mathrm{y}$ and $\tan \mathrm{x}$, and by $\operatorname{Sec}^{2}$:

$$\frac{\operatorname{Sec}^{2} \mathrm{y}}{\tan \mathrm{y}} , dy = -\frac{\operatorname{Sec}^{2} \mathrm{x}}{\tan \mathrm{x}} , dx$$

4. Integrate both sides:

   We now integrate both sides:

$$\int \frac{\operatorname{Sec}^{2} \mathrm{y}}{\tan \mathrm{y}} , dy = -\int \frac{\operatorname{Sec}^{2} \mathrm{x}}{\tan \mathrm{x}} , dx$$

On the left-hand side, the integral of $\frac{\operatorname{Sec}^{2} \mathrm{y}}{\tan \mathrm{y}}$ is $\ln \tan y$. Similarly, on the right-hand side, the integral of $\frac{\operatorname{Sec}^{2} \mathrm{x}}{\tan \mathrm{x}}$ is $-\ln \tan x$. This gives:

$$\ln \tan y = -\ln \tan x + \ln |C|$$

5. Simplify the equation:

   Combine the logarithmic terms:

$$\ln |\tan y| = \ln |\tan x| + \ln |C|$$

Using properties of logarithms:

$$\ln |\tan y \tan x| = \ln |C|$$

6. Final solution:

   Exponentiate both sides to remove the logarithms:

$$\tan x \tan y = C$$

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Key Formulas or Methods Used

• Separation of Variables: Rearranged the equation to separate $x$-terms and $y$-terms.
• Integration of Trigonometric Functions:
  • $\int \frac{\operatorname{Sec}^{2} \mathrm{y}}{\tan \mathrm{y}} , dy = \ln \tan y$
  • $\int \frac{\operatorname{Sec}^{2} \mathrm{x}}{\tan \mathrm{x}} , dx = -\ln \tan x$
• Logarithmic Properties: Used $\ln a + \ln b = \ln(ab)$ to combine the logarithmic terms.

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Summary of Steps

1. Start with the equation: $\operatorname{Sec}^{2} \mathrm{x} \tan \mathrm{y} , dx + \operatorname{Sec}^{2} \mathrm{y} \tan \mathrm{x} , dy = 0$.
2. Rearrange to: $\operatorname{Sec}^{2} \mathrm{y} \tan \mathrm{x} , dy = -\operatorname{Sec}^{2} \mathrm{x} \tan \mathrm{y} , dx$.
3. Separate the variables: $\frac{\operatorname{Sec}^{2} \mathrm{y}}{\tan \mathrm{y}} , dy = -\frac{\operatorname{Sec}^{2} \mathrm{x}}{\tan \mathrm{x}} , dx$.
4. Integrate both sides: $\ln \tan y = -\ln \tan x + \ln |C|$.
5. Simplify: $\ln |\tan y \tan x| = \ln |C|$.
6. Final result: $\tan x \tan y = C$.